Chapter 13 - Thermochemistry

+

Bonds are broken in the reactants :( +1370 kJ heat energy absorbed)

New bonds are formed in the products:(-1856 kJ heat energy released )

4541 CHEMISTRY Chapter 13

Form 5Chapter 13 : THERMOCHEMISTRY

Thermochemistry is the study of changes in heat energy during chemical reaction.Two types of chemical reactions :

1. Exothermic reaction2. Endothermic reaction

13.1 : ENERGY CHANGES IN CHEMICAL REACTIONS

1. Chemical energy is stored within the chemical bonds. During a chemical reaction, chemical bonds of the reactants are broken and new bonds in the products are formed.

When bonds are broken, heat energy is ………………………….

When bonds are formed, heat energy is ………………………….

2. Energy that is released or absorbed is in the form of ……………… energy.

3. The energy change (the difference between the energy of reactants and the products) in a chemical reaction is called heat of reaction, ∆H.

4. Heat of reaction, ∆H is the energy change when one mole of reactant reacts or when one mole of product is formed.

5. Two types of reactions that occur are :

Exothermic reaction Endothermic reaction

13.1.1 Exothermic Reaction

Example :

Chapter 13 Thermochemistry

2H2 + O2 → 2H2O , ∆H = - 486 kJ

∆H = Total energy content of products – total energy content of reactants

= Hproducts – Hreactants

2H2 + O2

Energy

Energy absorbed+1370 kJ

Energy released-1856 kJ

H = - 486 kJ

2H2O

The value of H is -486 kJ The heat released from bond formation is greater than heat absorbed for bond breaking.A negative sign for H shows that heat is released.

Products

Reactants t/b


:Energy change in the reaction :

Energy change :H = 1370 - 1856

= - 486 kJ

Energy Level Diagram

Energy level diagram shows the total energy content of the reactants compared to the products. Energy level diagram for exothermic reactions :

Energy

∆H negative (heat is released)

Total energy content of the products is less than total energy of the reactants.

Example of exothermic reactions :

Chemical equation

Neutralisation 2KOH + H2SO4 → K2SO4 + 2 H2O

Reaction between acids and metals Mg + 2HCl → MgCl2 + H2

Reaction between acids and carbonate Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2

Combustion of alcohol C2H5OH + 3O2 → 2CO2 + 3H2O

Dissolving sodium hydroxide in water NaOH(s) → Na+ (aq) + OH- (aq)


Interpretation :

The quantity of energy absorbed for bonds breaking between

hydrogen atoms and oxygen atoms are ……………….. than

energy released for the formation of bonds between hydrogen

and oxygen atoms.

…………. bonds are broken and ………….. bonds are formed.

Heat is released to the surrounding temperature increases.

The sign of ∆H is …………………….

Energy change : Chemical energy → Heat energy

Total energy content are decreased. Total energy content of the

product is lower /less than total energy of the reactants.

lower

Weak strong

negative

Interpretation :The quantity of energy absorbed for bonds breaking is ……… than energy released from the formation of bonds.………... bonds are broken and ………... bonds are formed.Heat is absorbed from the surrounding temperature decreasesThe sign of ∆H is …………………….Energy change : Heat energy → Chemical energy Total energy content are increased. Total energy content of the product is more than total energy of the reactants.

Energy absorbed(+ve)

Reactants

Energy released( -ve)

Hpositive

Products

Productsb

Reactants


Adding water to concentrated acid. H2SO4(aq) → 2H+ (aq) + SO42-(aq)

13.1.2 Endothermic Reaction

Energy Change in Endothermic Reaction:

Energy

Energy Level Diagram

Energy level diagram shows the total energy content of the reactants compared to the products . Energy level diagram for exothermic reactions :

Energy

∆H positive (heat is absorbed)

Total energy content of the products is greater /more than total energy of the reactants.

Example of endothermic reactions :

Chemical equation

Decomposition of nitrate and carbonate salt when heated CaCO3 CaO + CO2

Decomposition of hydrated salt to anhydrous salt and water when heated

CuSO4.5H2O(s) CuSO4(s) + 5H2O(l) (blue) (white)

Dissolving ammonium salts/nitrate salts in water

H2O NH4Cl(s) NH4

+ (aq) + Cl- (aq)


Strong

higher

weak

positive

Energy

Reactants

Products

∆H = -ve

Energy

Reactants

Products

∆H = +ve

Energy

HCl + NaOH

NaCl + H2O

∆H = - 57 kJ mol-1

NH4NO3

NH4+ + NO3-

∆H = + 26 kJ mol-1


Activity 1:-

1 Complete the following table to compare and contrast between Exothermic and Endothermic reactions.

Exothermic Endothermic

Absorb /release heat Release heat Absorb heat

Temperature of surrounding change

Increase Decrease

Total energy content Reactants products Products reactants

Energy level diagram

Value of ∆H Negative Positive

2 Construct energy level diagram for the following thermochemical equations :

(i) HCl + NaOH → NaCl + H2O ∆ H = -57 kJ

(ii) NH4NO3(s) → NH4+(aq) + NO3 -(aq) ∆ H = +26 kJ


Energy

H2O

H2O


13.2 : APPLICATION OF KNOWLEDGE OF EXOTHERMIC AND ENDOTHERMIC REACTIONS IN EVERYDAY LIFE

(a) Hot packs

Contain chemicals that released heat [exothermic reaction]It is a plastic bag containing separate compartments of water and anhydrous calcium chloride.The anhydrous calcium chloride dissolve in water to release heat; the temperature increase. CaCl2(s) Ca2+(aq) + 2Cl-(aq) ∆H = - 83 kJ

Other substances that can be used in a hot pack are anhydrous magnesium sulphate, anhydrous copper(II) sulphate and calcium oxide.A reusable hot pack uses supersaturated solution of sodium ethanoate crystallization and resolution.

(b) Cold packs

Contain chemicals that absorbed heat [endothermic reaction]It is a plastic bag containing separate compartments of water and solid ammonium nitrate.The solid ammonium nitrate dissolve in water and absorbed heat from surrounding;the temperature decrease.

NH4NO3(s) NH4

+(aq) + NO3 - (aq) ∆H = + 83 kJ

Other substances that can be used in a cold pack are ammonium chloride, potassium nitrate and sodium thiosulphate.



4.3 : DETERMIINE THE HEAT OF REACTION (∆H)

1. Heat of Reaction, ∆H

the energy change when one mole of reactant reactsor

the energy change when one mole of product is formed.

2. Four types of heat of reaction discussed in this chapter are :

Heat of reaction

Definition Example

1Heat of Precipitation

Heat change/ Heat energy released when 1 mol of precipitate is formed.

Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq),

∆H = -50.4 kJ

Ionic equation :

Pb2+ + SO42- → PbSO4

50.4 kJ heat energy is released when 1 mol of lead(II) ions reacted with 1 mol of sulphate ions to form 1 mol of lead (II) sulphate.

2Heat of Displacement

Heat change/ Heat energy released when 1 mol of metal is displaced from its salt solution.

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), ∆H = -217 kJ

Ionic equation :

Cu2+ + Zn → Zn2+ + Cu

217 kJ heat energy is released when 1 mol of copper is displaced from copper(II) sulphate solution by zinc.

3Heat of Neutralisation

Heat change / Heat energy released when 1 mol of water is formed from neutralisation of acid with an alkali.

KOH(aq) + HNO3 (aq)→ KNO3(aq) + H2O(l), ∆H = -57 kJ

Ionic equation :

H+ + OH- → H2O

57 kJ heat energy is released when 1 mol of water formed from neutralization of potassium hydroxide with nitric acid.

4Heat of Combustion

Heat change / Heat energy released when 1 mol of fuel is burnt completely in excess oxygen.

C2H5OH + 3O2 → 2CO2 + 3H2O, ∆H = -1366 kJ

1366 kJ heat energy is released when 1 mol of ethanol is burnt completely in excess oxygen .


Heat change, H = mc (J)

Heat change = mc m = mass of the solution in gram c = specific heat capacity of solution in J g-1 oC-1 = temperature change in ºC

Hn

Hn


3. Method of calculating ∆H

Quantity of heat change in a substance depends on :

Mass of substance, m (g)

Specific heat capacity of a substance, c ( J g-1 oC-1 )

Temperature change, (oC)

As the chemical reaction occurs in an aqueous solution, these assumptions are made during the calculation of heat of reaction :

Density of aqueous solution = Density of water = 1 g cm-3

1 cm3 of aqueous solution has a mass of 1 g

250 cm3 of aqueous solution has a mass of 250 g

x cm3 of aqueous solution has a mass of x g

Specific heat capacity of solution, c = Specific heat capacity of water = 4.2 J g-1 oC-

1

No heat lost to the surroundings during reaction, all heat released in an exothermic reaction is absorbed into the reaction mixture.

Heat change, H

The heat change in a reaction can be calculated the following formula :

Heat of reaction , ∆H

Heat of reaction ( ∆H) is the energy change when

one mole of reactant reactsor

one mole of product is formed.

Calculation :

If n mol of reactant/product absorbs/releases H J of heat energy,

1 mol of reactant/product absorbs/releases J mol -1

∆H (heat of reaction) = +/- J

Note : i. The sign of ∆H is negative for exothermic reaction ( temperature increases)ii. The sign of ∆H is positive for endothermic reaction (temperature decreases)


x = number of moles of reactant/product


iii. The unit for heat of reaction is kJ mol-1.iv. When the amount of heat is presented in ∆H, it is written ∆H = +/- ……kJ mol-1.

Example 1 : (Heat of Precipitation)

60 cm 3 of 0.025 mol dm -3 silver nitrate solution reacts with 60 cm 3 of 0.025 mol dm -3 potassium bromide solution at a temperature of 29 oC. A yellow precipitate was formed and the highest temperature reached is 32 oC. Determine the heat of reaction, ∆H and draw the energy level diagram for this reaction.

Solution :

Steps Calculation

S1: Determine the mass of the solution, m (Density of aqueous solution = 1 g cm-3)

Mass of the solution, m = ( 60 cm3 + 60 cm3 ) × 1 g cm-3

m = 120 g

S2 : Determine the temperature change, θTemperature change, θ = 32 – 29

= 3 oC

S3 : Determine the energy change/ heat released, H

(Specific heat capacity of solution = 4.2 J g-1 oC-1)

Heat released, H = mcθ= 120 × 4.2 × 3

= 1512 J

S4 : Determine the number of moles of silver bromide precipitated, n Number of moles of Ag+ = 0.025 ×

601000

= 0.0015 mol

Number of moles of Br- = 0.025 ×

601000

= 0.0015 mol

AgNO3 + KBr AgBr + KNO3

orAg+ + Br- AgBr

From the ionic equation :1 mol of Ag+ ions reacts with 1 mol of Br- ions to form 1 mol of AgBr0.0015 mol of Ag+ ions reacts with 0.0015 mol Br- ions to form 0.0015 mol of AgBr.

Number of mole of AgBr, n = 0.0015 mol


Hn

15120.0015

Energy

AgNO3 + KBr

AgBr + KNO3

∆H = - 100.8 kJ mol-1

Energy

Ag+ + Br-

AgBr

∆H = - 100.8 kJ mol-1


S5 : Determine the heat of reaction, ∆H ∆H = -

= -

= - 100800 J mol-1

= - 1008 kJ mol-1

Energy Level Diagram :

or


∆H is negative because temperature increased / heat is released to the surrounding (Exothermic reaction).

From chemical equation

From ionic equation


Example 2 : (Heat of Displacement)

Excess zinc, Zn powder is added to 50 cm3 of 0.05 mol dm-3 copper(II) nitrate, Cu(NO3)2 solution. The temperature of the reaction mixture rises by 2.62 oC. Calculate the heat of displacement, ∆H of copper, Cu from its salt solution and draw the energy level diagram for this reaction.

Solution :

Steps Calculation

S1: Determine the mass of the solution, m (Density of aqueous solution = 1 g cm-3)

Mass of the solution, m = 50 cm3 × 1 g cm-3

m = 50 g

S2 : Determine the temperature change, θ Temperature change, θ = 2.62 oC

S3 : Determine the heat released, H(Specific heat capacity of solution = 4.2 J g-1 oC-1)

Heat released, H = mcθ= 50 × 4.2 × 2.62

= 550.2 J

S4 : Determine the number of moles of copper(II) nitrate, n Num. of moles of Cu(NO3)2 / Cu2+ = 0.05 ×

501000

= 0.0025 mol

Cu(NO3)2 + Zn Cu + Zn(NO3)2

From the equation :1 mol of Cu(NO3)2 form 1 mol of Cu0.0025 mol of Cu(NO3)2 form 0.0025 mol of Cu

Number of mole of Cu displaced, n = 0.0025 mol


Hn

550.20.0025

Energy

Cu(NO3)2 + Zn

Cu + Zn(NO3)2

∆H = - 220.08 kJ mol-1

Energy

Cu2+ + Zn

Cu + Zn2+

∆H = - 220.08 kJ mol-1



= -

= - 220080 J mol-1

= - 220.08 kJ mol-1


or


∆H is negative because temperature increased/rised. Heat is released to the surrounding (Exothermic reaction)

From chemical equation

From ionic equation

Butanol

Thermometer

Copper can

250 cm3 of water

Wooden block

Wind shield


Example 3 : (Heat of Combustion)

Diagram below shows the set-up of apparatus for an experiment to determine the heat of combustion of butanol.

Result :

Volume of water in copper can, m = 250 cm3

Initial temperature of water, T1 = 28.0 0CThe highest temperature of water, T2 = 59.5 0C

Initial mass of spirit lamp contains butanol = 175.20 g

Final mass of spirit lamp = 174.10 g

Calculate the heat of combustion, ∆H of butanol, C4H9OH and draw the energy level diagram for this reaction.

Solution :

Steps Calculation


Mr of C4H9OH = 4(12) + 10(1) +16= 74

Hn

120750.015

Energy

C4H9OH + 6O2

4CO2 + 5H2O

∆H = - 805 kJ


S1: Determine the mass of the water, m (Density of aqueous solution = 1 g cm-3)

Mass of the solution, m = 250 cm3 × 1 g cm-3

m = 250 g

S2 : Determine the temperature change, θTemperature change, θ = 59.5 – 28.0

= 11.5 oC

S3 : Determine the heat released, H(Specific heat capacity of solution = 4.2 J g-1 oC-1)

Heat released, H = mcθ= 250 × 4.2 × 11.5

= 12075 J

S4 : Determine the number of moles of butanol is burnt, n

Mass of C4H9OH = 175.20 – 174.10= 1.10 g

Num. of moles of C4H9OH =

1. 1074

= 0.015 mol


= -

= - 805000 J mol-1

= - 805 kJ mol-1

Chemical equation :

C4H9OH + 6O2 → 4CO2 + 5H2O



∆H is negative because temperature increased/rised. Heat is released to the surrounding (Exothermic reaction)



Metal powder

Salt solution

Thermometer

Polystyrene cup

Acid Alkali Alkali

Acid


13.4 ACTIVITY / EXPERIMENT TO DETERMINE HEAT OF REACTION

Heat of Reaction

Procedure m c θ n

Heat of displacement

Set up of apparatus :

Procedure :1. Measure the volume of salt* solution using a measuring

cylinder.2. Pour the salt solution into a polystyrene cup.3. Determine the initial temperature (T1).4. Quickly and carefully, excess metal** powder is added

into the solution.5. The mixture is stirred with a thermometer and the

highest temperature reached is recorded (T2).

Volume of salt solution

Specific heat capacity of water

θ = T2 – T1

Number of moles of metal displaced[calculated from balanced equation of displacement reaction]

Heat of neutralisation


Procedure :1. Measure the volume of acid* and alkali** using measuring

cylinders.2. Pour the solutions into different polystyrene cups.3. Record the initial temperature of acid & alkali (Ta and Tb).4. Quickly and carefully, acid is poured into the alkali.5. The mixture is stirred with a thermometer and the

highest temperature reached is recorded (T2).

Total volume of acid and alkali


T1 = Ta + Tb

2

θ = T2 – T1

Number of moles of water formed.[calculated from balanced equation of neutralisation reaction]


Solution A

Solution B

Fuel /alcohol

Thermometer

Copper can

Water


Heat of precipitation


Procedure :[Use two aqueous solutions that contain the cation and the anion of the insoluble salt]. Procedure is the same as above.

Total volume of both aqueous salt solutions


θ = T2 – T1

Number of moles of precipitate [calculated from balanced equation of reaction]

Heat of combustion


Procedure :1. Determine the mass of fuel used by measuring the weight

of the lamp with the fuel before burning, m1 and after burning, m2.

2. The heat released during burning is used to raise the temperature of water in the copper can [determine the initial temperature of water, T1 and the highest temperature, T2]

Volume of water in the copper can


θ = T2 – T1

m1 –m2

RMM of fuel


H+ + OH- → H2O

∆H = - 57 kJ mol-1


13.4.1 Comparison of Heat of Neutralisation

(a) Strong a cid and strong alkali :

All neutralisation process can be represented by the following equation :

H+ + OH- → H2O, ∆H = - 57 kJ mol–1

1 mol of hydrogen ions react with 1 mol of hydroxide ions to form 1mol of water to release 57 kJ of heat energy.

Heat of neutralisation for KOH/ NaOH with HCl and HNO3 is the same because all these reactions form 1 mol of H2O.

HCl + KOH → KCl + H2O HCl + NaOH → NaCl + H2OHNO3 + KOH → KNO3 + H2O

HNO3 + NaOH → NaNO3 + H2O

Neutralisation of NaOH with H2SO4 (diprotic acid)

2NaOH + H2SO4 → Na2SO4 + 2H2O

2 mol of OH- reacts with 2 mol of H+ to form 2 mol of H2O.

Heat released is 2 × 57 kJ, that is, 114 kJ, not 57 kJ.

Heat of neutralisation of sulphuric acid with sodium hydroxide remains at -57 kJ mol-1 because the definition for heat of neutralisation is in terms of formation of 1 mol of water, not 2 mol of water.

(b) Weak a cid and strong alkali :

Magnitude of heat of neutralisation for a weak acid with a strong alkali is less than 57 kJ mol-1.

NaOH + CH3COOH → CH3COONa + H2O, ∆H = -55 kJ mol-1

NaOH + HCN → NaCN + H2O, ∆H = -12 kJ mol-1

Explanation :

Weak acids ionise partially in water to produce hydrogen ions in low concentration.

CH3COOH CH3COO- + H+

Some of the acid particles still remain in the form of molecules.

Heat energy is absorbed to break the bonds in the molecules of the weak acid that have not been ionised, so that they ionise completely.

Part of the heat that is released is used to break the bonds in the molecules of the weak acid that has not been ionised.


ethanoic acid

hydrocyanic acid

17

Number of mole of HCl = Number of mole of KOH

Number of mole of H2O = Number of mole of HClNumber of mole of H2O = Number of mole of KOH

KOH + HCl → KCl + H2O

Number of mole of HCl = Number of mole of H+Number of mole of KOH = Number of mole of OH-


Example 1 :

Experiment Reactants Temperature change

I50 cm3 of 2.0 mol dm-3 hydrochloric acid is added to 50 cm3 of 2.0 mol dm-3 potassium hydroxide solution.

rises by 13 oC

II300 cm3 of 2.0 mol dm -3 hydrochloric is added to 300 cm3 of 2.0 mol dm -3 potassium hydroxide solution.

What is the temperature change in Experiment II?

☺ Solution :Experiment I :

Heat released, H = mcθ= (50 + 50) × 4.2 × 13= 5460 J

Number of mole of HCl / H+ = MV1000

= 2.0 × 50 1000

= 0.1 mol

Heat of neutralisation, ∆H =

54600 .1

= 54600 J

Heat of neutralisation, ∆H for Exp. I = Heat of neutralisation, ∆H for Exp. II

Experiment II :

Heat of neutralisation, ∆H = 54600 J

Number of mole of HCl / H+ = MV1000

= 2.0 × 300 1000

= 0.6 mol

Heat released, H = 54600 × 0.6= 32760 J


Calculation guide :

(i) If the experiment is repeated by changing the volume without changing the concentration, change in temperature is the same.

(ii) If the experiment is repeated by changing the concentration of the solution by x times without changing the volume, the temperature change is x times.

18

∆H= Heat of precipitation of lead(II) carbonate

H = Heat change = mc

n= Number of moles of lead(II) carbonate


Heat released, H = mcθmcθ = 32760

(300 + 300) × 4.2 × θ = 32760

θ =

327602520

= 13 0C

Example 2 :

Experiment Reactants Temperature change

P50 cm3 of 0.2 mol dm-3 lead(II) nitrate solution is added to 50 cm3 of 0.2 mol dm-3 sodium carbonate solution.

rises by 2.4 oC

Q50 cm3 of 0.6 mol dm-3 lead(II) nitrate solution is added to 50 cm3 of 0.6 mol dm-3 sodium carbonate solution.

What is the temperature change in Experiment Q?

Solution :

∆H = Hn

Ionic equation for both experiments :

Pb2+ + CO32- PbCO3

Experiment P :

∆H = 100 × 4.2 × 2.4 0.01

= 100800 J

Experiment Q :

100800 = 100 × 4.2 × 0.03

= 7.2 oC

[ The temperature changes 3 times more than Exp. P! ]

.


19

Heat of combustion of alcohol (kJ mol -1)

Number of carbon atoms per molecule12345

4000

3000

2000

1000

Thermometer

Tripod stand

Copper container

Spirit lamp

Wooden block

Alcohol*

Water

Wind shield


13.4.2 Comparison between Heat of Combustion of Various Fuel

1. The higher the number of carbon and hydrogen atoms per molecule, the higher the heat energy released by the combustion of 1 mol of fuel.

The heat of combustion of alcohols increase with the increasing of number of carbon atoms and hydrogen atoms in the molecules.

More heat released when more carbon atoms become carbon dioxide molecules and hydrogen atoms become water molecules.

Example :

Diagram below shows the set-up of apparatus for an experiment to compare the heat of combustion of methanol and ethanol. 200 cm3 of water is poured into copper container in this experiment.


20

20

25

15

20

25

15

60

65

55

60

65

55

150.50 151.35

149.37 150.50


Table below shows the results obtained from Experiment I; to determine the heat of combustion of methanol and Experiment II; to determine the heat of combustion of ethanol.

Experiment I Experiment II

Initial temperature of water : …………………… Initial temperature of water : .........................

Highest temperature of water : ...........................

Highest temperature of water : ..........................


Initial mass of spirit lamp and methanol : ......... g

Initial mass of spirit lamp and ethanol : …......... g

Final mass of spirit lamp and methanol : ……... g Final mass of spirit lamp and ethanol : .............. g


21


(a) Write the readings of the temperature and the mass of spirit lamp and alcohol in the spaces provided.

(b) Based on the experiment above, complete the table below.

Name of variables Action to be taken

(i) Manipulated variable :

………………..........……………………..

………………..........……………………..

………………..........……………………..

(i)

………………..........……………………..

………………..........……………………..

………………..........……………………..

(ii) Responding variable :

………………..........……………………..

………………..........……………………..

………………..........……………………..

(ii) What to observe in the responding variable :

………………..........……………………..

………………..........……………………..

………………..........……………………..

(iii) Fixed variable :

………………..........……………………..

………………..........……………………..

………………..........……………………..

(iii) The way to maintain the controlled variable :

………………..........……………………..

………………..........……………………..

………………..........……………………..

(c) (i) Calculate the energy change, H in the Experiment I and II.[ Energy change = mc, specific heat capacity of water, c = 4.2 J g-1 oC-1 ]


(ii) Calculate the number of moles of methanol and ethanol burnt in this experiment.[ Molar mass of methanol = 32 g mol-1 ; Molar mass of ethanol = 46 g mol-1 ]

Methanol Ethanol


22


(iii) Calculate the heat of combustion of methanol in this experiment.

Heat of combustion of methanol Heat of combustion of ethanol

(d) Compare to the heat combustion of methanol and ethanol. Explain why.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

2. Fuel Value Fuel value is the amount of heat released when 1 g of fuel burns completely,

the unit is kJ g-1

A fuel with high fuel value can supply more energy.

Example :

Type of Fuel Fuel Value / kJ g-1

Methanol 23Charcoal 35Crude oil 45Kerosene 37

Petrol 34Natural gas 50

Aspects to be considered when choosing a fuel in industry :

(a) Fuel value of the fuel.(b) Cost of energy / cost of fuel.(c) Availability and sources of the fuel.(d) Effect of the fuel to the environment.

4.4 OTHER SOURCES OF ENERGY

World’s major sources of energy are fossil fuels such as coal, petroleum and natural gas are non-renewable source of energy, eventually they will be used up.

Other sources of energy are the sun, biomass, water and radioactive substances.

END OF CHAPTER 4


23

Chapter 13 - Thermochemistry

Documents

energy level diagram

highest temperature reached

total energy content

4co2 5h2o

specific heat capacity

temperature change rises

ag ions reacts

highest temperature